Problem: Perform the row operation, $R_1+6R_3\rightarrow R_1$, on the following matrix. $\left[\begin{array} {ccc} 0 & -2 & 3 & 1 \\ -9 & 9 & -4 & 2 \\ 3 & -2 & -3 & 2 \end{array} \right] $
Background There are three basic row operations that can be performed on matrices. $R_i \leftrightarrow R_j$. This symbol tells us to interchange rows $i$ and $j$. $cR_i \rightarrow R_i$. This symbol tells us to multiply a row $i$ by a constant $c$. $R_i + cR_j \rightarrow R_i$. This symbol tells us to add $c$ times row $j$ to row $i$. Finding the new row to be used For the given matrix, $R_1$ and $R_3$ are given below. $R_1=\left[\begin{array} {ccc} 0 & -2 & 3 & 1 \end{array} \right] ~~~~~ R_3=\left[\begin{array} {ccc} 3 & -2 & -3 & 2 \end{array} \right]$ We are asked to perform the row operation, $R_1+6R_3\rightarrow R_1$. Therefore, we must add $6R_3$ to $R_1$. $\begin{aligned}R_1+6R_3&= \left[\begin{array} {ccc} 0 & -2 & 3 & 1 \end{array} \right] + 6\left[\begin{array} {ccc} 3 & -2 & -3 & 2 \end{array} \right] \\\\&=\left[\begin{array} {ccc} 18 & -14 & -15 & 13 \end{array} \right]\end{aligned}$ Substituting the row Now, we must substitute row $R_1$ with $R_1+6R_3$. $\left[\begin{array} {ccc} {0} & {-2} & {3} & {1} \\ -9 & 9 & -4 & 2 \\ 3 & -2 & -3 & 2 \end{array} \right] \xrightarrow{R_1+6R_3\rightarrow R_1} \left[\begin{array} {ccc} {18} & {-14} & {-15} & {13} \\ -9 & 9 & -4 & 2 \\ 3 & -2 & -3 & 2 \end{array} \right] $ Summary Our resultant matrix is the following. $\left[\begin{array} {ccc} 18 & -14 & -15 & 13 \\ -9 & 9 & -4 & 2 \\ 3 & -2 & -3 & 2 \end{array} \right]$